The function $\mathrm{f}$ is defined by $\mathrm{f} : x \mapsto 1 - (x + 3)^2$, \(x \in \mathbb{

Question

The function $\mathrm{f}$ is defined by $\mathrm{f} : x \mapsto 1 - (x + 3)^2$, $x \in \mathbb{R}$, $x \ge -2$. Show that the inverse function of $\mathrm{f}$ exists. Find $\mathrm{f}^{-1}(x)$ and state its domain.[4] On the same diagram, sketch the graphs of $y = \mathrm{f}(x)$, $y = \mathrm{f}^{-1}(x)$ and $y = \mathrm{f}^{-1}\mathrm{f}(x)$, showing clearly the relationship between the graphs.[3] Hence, find the exact solution of $\mathrm{f}^{-1}\mathrm{f}(x) \le \mathrm{f}(x)$.[3]

Accepted Answer

(i) $\mathrm f^{-1}(x)=-3+\sqrt{1-x}$, domain $(-\infty,0]$ (ii) $y=\mathrm f^{-1}(x)$ is the reflection of $y=\mathrm f(x)$ in $y=x$, and $y=\mathrm f^{-1}\mathrm f(x)=x$ for $x\ge-2$ (iii) $-2\le x\le\dfrac{-7+\sqrt{17}}2$

2011 NJC P1 Q7

2011 NJC P1 Q7

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